Question

Find the values of p for which the quadratic equation $\left(p+1\right)x^2-6\left(p+1\right)x+3\left(p+9\right)=0,p\ne -1$ has equal roots. Hence, find the roots of the equation. [CBSE 2015]

Answer 1

The given equation is $\left(p+1\right)x^2-6\left(p+1\right)x+3\left(p+9\right)=0$.

This is of the form $a{x}^2+bx+c=0$, where a = p +1, b = −6(p + 1) and c = 3(p + 9).

$$\begin{gathered} \therefore D=b^2-4ac \\ ={\left[-6\left(p+1\right)\right]}^2-4\times \left(p+1\right)\times 3\left(p+9\right) \\ =12\left(p+1\right)\left[3\left(p+1\right)-\left(p+9\right)\right] \\ =12\left(p+1\right)\left(2p-6\right) \end{gathered}$$

The given equation will have real and equal roots if D = 0.

$$\begin{gathered} \therefore 12\left(p+1\right)\left(2p-6\right)=0 \\ \Rightarrow p+1=0\mathrm{or}2p-6=0 \\ \Rightarrow p=-1\mathrm{or}p=3 \end{gathered}$$

But, $p\ne -1$ (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4x^2-24x+36=0$.

$$\begin{gathered} 4x^2-24x+36=0 \\ \Rightarrow 4\left(x^2-6x+9\right)=0 \\ \Rightarrow {\left(x-3\right)}^2=0 \\ \Rightarrow x-3=0 \end{gathered}$$
$\Rightarrow x=3$

Hence, 3 is the repeated root of this equation.